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Let $\alpha$ be a root of $x^4+4kx+1=0$ where $k$ is an integer. Is $\Bbb Q[\alpha]=\{a_0+a_1\alpha+a_2\alpha^2+a_3\alpha^3; a_i\in\Bbb Q\}$ a field?

I find it is quite hard to see $\Bbb Q[\alpha]$ is closed under the division operator. When I write $1/(a_0+a_1\alpha+a_2\alpha^2+a_3\alpha^3)=b_0+b_1\alpha+b_2\alpha^2+b_3\alpha^3$, I find it is not easy to find $b_i$.

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  • $\begingroup$ see the questions I asked in comments there math.stackexchange.com/questions/1718969/… to see why $\mathbb{Q}(c_1,\ldots,c_n)$ is a finite dimensional $\mathbb{Q}$ vector space when the $c_i$ are algebraic $\endgroup$ – reuns Apr 3 '16 at 3:46
  • $\begingroup$ If you know a little bit of field theory and polynomial theory in abstract algebra, you can look at the answer below. If not, I think there is no easy elementary approach available. $\endgroup$ – Vim Apr 3 '16 at 3:53
  • $\begingroup$ @Vim : the one I linked to. if $b \in L$ where $L = \mathbb{Q}[c_1,\ldots, c_n]$ where the $c_i$ are the roots of some irreducible polynomial of $\mathbb{Q}[x]$, then $b^k \in L$ for every $k \in \mathbb{Z}$. hence that vector space $L$ is a field $\endgroup$ – reuns Apr 3 '16 at 3:57
  • $\begingroup$ @user1952009 still requires abstract algebra. I assume OP has only the knowledge of LA but not AA. $\endgroup$ – Vim Apr 3 '16 at 4:04
  • $\begingroup$ @Vim : I don't agree, you don't need any AA knowledge, just by induction that every element $b$ in the vector space $L$ has a minimal polynomial, hence using the minimal polynomial for the $c_i$, the powers of $b$ are also in $L$ and also have a minimal polynomial. just induction, no need of complicated AA knowledge. and that proof is exactly the same as proving that the algebraic numbers are a field : it is a basic proof for constructing the rest of AA $\endgroup$ – reuns Apr 3 '16 at 4:11
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You ought to look at it in the intended way: $\Bbb Q[\alpha]\cong \Bbb Q[x]/(x^4+4kx +1)$. If you show the polynomial is irreducible over $\Bbb Q$, then the quotient is a field. It's highly unlikely that anyone intends you to find inverses directly.

By the rational root test, it certainly has no linear factors. All that remains is to eliminate the possibility of two quadratic polynomial factors.

At the worst , you can brute force that question. Attempt to solve $(x^2+ax+b)(x^2+cx+d)=x^4+4kx+1$ for a factorization.

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